The inequality is $ | \left< x,y \right>| \le \|x\|\|y\|. $ with standard inner product definition. One neat trick to prove this is using an auxilary parameter $t,$ and expanding $$ \| x+ty \| =
\left< x+ty,x+ty \right> = \|x\| + 2 \left< x,y \right>t +\|y\|t^2.$$ We know, this being a square, is greater or equal to zero. Therefore, the discriminant of the polynomial in $t$ is less or equal to zero. Which is $\left< x,y \right>^2 - \|x\|\|y\| \le 0.$ Substituting the values for $x$ and $y$ will do the job.